Mixing
How can I find the hazard rate of a time-dependent accelerated failure time model
$begingroup$
Suppose I have the following accelerated failure time model:
$t = exp(-beta t)u$
where
the c.d.f of $u$, $F_u(x) = dfrac{x}{1+x}$.
I tried the following way:
$u = texp(beta t) = Psi(t)$, or $t = Psi^{-1}(u)$
$F_t(x)=Pr(t leq x) = Pr(Psi^{-1}(u) leq x) = Pr(u leq Psi(x))$
$F_t(x) = F_u(Psi(x)) = dfrac{Psi(x)}{1+Psi(x)}=dfrac{xexp(beta x)}{1+xexp(beta x)}$
The p.d.f of $t$ is:
$f_t(x)=dfrac{(1+beta x)exp(beta x)}{(1+xexp(beta x))^2}$
The problem is, when $beta$ is a negative parameter, $1+beta x <0$ is possible, and we may end up with a negative p.d.f!
I know there must be something wrong with my procedure, just don't know where.
stochastic-processes
asked Jan 28 at 15:49
skyindeerskyindeer
386
$endgroup$
add a comment |
$begingroup$
Suppose I have the following accelerated failure time model:
$t = exp(-beta t)u$
where
the c.d.f of $u$, $F_u(x) = dfrac{x}{1+x}$.
I tried the following way:
$u = texp(beta t) = Psi(t)$, or $t = Psi^{-1}(u)$
$F_t(x)=Pr(t leq x) = Pr(Psi^{-1}(u) leq x) = Pr(u leq Psi(x))$
$F_t(x) = F_u(Psi(x)) = dfrac{Psi(x)}{1+Psi(x)}=dfrac{xexp(beta x)}{1+xexp(beta x)}$
The p.d.f of $t$ is:
$f_t(x)=dfrac{(1+beta x)exp(beta x)}{(1+xexp(beta x))^2}$
The problem is, when $beta$ is a negative parameter, $1+beta x <0$ is possible, and we may end up with a negative p.d.f!
I know there must be something wrong with my procedure, just don't know where.
stochastic-processes
asked Jan 28 at 15:49
skyindeerskyindeer
386
$endgroup$
add a comment |
$begingroup$
Suppose I have the following accelerated failure time model:
$t = exp(-beta t)u$
where
the c.d.f of $u$, $F_u(x) = dfrac{x}{1+x}$.
I tried the following way:
$u = texp(beta t) = Psi(t)$, or $t = Psi^{-1}(u)$
$F_t(x)=Pr(t leq x) = Pr(Psi^{-1}(u) leq x) = Pr(u leq Psi(x))$
$F_t(x) = F_u(Psi(x)) = dfrac{Psi(x)}{1+Psi(x)}=dfrac{xexp(beta x)}{1+xexp(beta x)}$
The p.d.f of $t$ is:
$f_t(x)=dfrac{(1+beta x)exp(beta x)}{(1+xexp(beta x))^2}$
The problem is, when $beta$ is a negative parameter, $1+beta x <0$ is possible, and we may end up with a negative p.d.f!
I know there must be something wrong with my procedure, just don't know where.
stochastic-processes
asked Jan 28 at 15:49
skyindeerskyindeer
386
$endgroup$
Suppose I have the following accelerated failure time model:
$t = exp(-beta t)u$
where
the c.d.f of $u$, $F_u(x) = dfrac{x}{1+x}$.
I tried the following way:
$u = texp(beta t) = Psi(t)$, or $t = Psi^{-1}(u)$
$F_t(x)=Pr(t leq x) = Pr(Psi^{-1}(u) leq x) = Pr(u leq Psi(x))$
$F_t(x) = F_u(Psi(x)) = dfrac{Psi(x)}{1+Psi(x)}=dfrac{xexp(beta x)}{1+xexp(beta x)}$
The p.d.f of $t$ is:
$f_t(x)=dfrac{(1+beta x)exp(beta x)}{(1+xexp(beta x))^2}$
The problem is, when $beta$ is a negative parameter, $1+beta x <0$ is possible, and we may end up with a negative p.d.f!
I know there must be something wrong with my procedure, just don't know where.
stochastic-processes
stochastic-processes
asked Jan 28 at 15:49
skyindeerskyindeer
386
asked Jan 28 at 15:49
skyindeerskyindeer
386
asked Jan 28 at 15:49
skyindeerskyindeer
386
asked Jan 28 at 15:49
skyindeerskyindeer
386
asked Jan 28 at 15:49
skyindeerskyindeer
386
386
add a comment |
add a comment |
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